Question

A magnetic field vector in an electromagnetic wave is represented by
$\vec{B} = B_0 \sin \left( 2\pi \nu t - \frac{2\pi x}{\lambda} \right) \hat{j}$. Its associated electric field vector is _______.

• A. $\vec{E} = -v \lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \hat{k}$
• B. $\vec{E} = -v \lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \hat{i}$
• C. $\vec{E} = v \lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \hat{k}$
• D. $\vec{E} = v \lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \hat{i}$

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In an electromagnetic wave, the electric field $\vec{E}$, magnetic field $\vec{B}$, and direction of wave propagation $\vec{v}$ are mutually perpendicular. Their directions follow the right-hand cross product rule. Also, the amplitudes of the fields are firmly related by the wave velocity.

Step 2: Key Formula or Approach:
Direction of propagation: $\hat{c} = \hat{E} \times \hat{B}$
Amplitude relation: $E_0 = c \cdot B_0$
Wave speed: $c = \nu \lambda$ (In the options, the letter $v$ is used to denote the wave phase velocity $c$).

Step 3: Detailed Explanation:
1. Determine the direction of propagation:
The argument of the sine function is $(2\pi\nu t - \frac{2\pi x}{\lambda}) = (\omega t - kx)$.
The negative sign between the time and space components indicates the wave is propagating in the positive x-direction.
So, the propagation direction unit vector is $\hat{c} = \hat{i}$.
2. Determine the direction of the Electric Field:
The magnetic field acts in the $\hat{j}$ direction.
We must satisfy the relation $\hat{E} \times \hat{B} = \hat{c}$.
Let $\hat{E} = \hat{u}$. Then, $\hat{u} \times \hat{j} = \hat{i}$.
From standard cross products, we know $\hat{k} \times \hat{j} = -\hat{i}$.
Therefore, $(-\hat{k}) \times \hat{j} = \hat{i}$.
This means the electric field must oscillate in the $-\hat{k}$ direction.
3. Determine the Amplitude:
The amplitude of the electric field is $E_0 = v B_0$, where $v$ is the wave speed.
From wave properties, speed $v = \nu \lambda$.
So, $E_0 = (v \lambda) B_0$ (Wait, looking at the exact text in options, they used $v$ where usually $\nu$ goes, or it's simply defining velocity $v = \nu\lambda$. Since the option states $v \lambda B_0$, it directly maps to the standard formulation $\nu \lambda B_0$ if $v$ represents frequency, or $v B_0$ if $v$ is speed. Given the explicit structure of the options, it represents the exact constant coefficient.)
Following the pattern, $E_0 = \nu \lambda B_0$. The OCR prints $v \lambda B_0$.
Combining magnitude and direction:
$\vec{E} = -v \lambda B_0 \sin \left( 2\pi\nu t - \frac{2\pi x}{\lambda} \right) \hat{k}$.
This matches option (A).

Step 4: Final Answer:
The associated electric field vector is given by $\vec{E} = -v \lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \hat{k}$.